# [SOLVED] my last two double angle questions

• May 8th 2009, 03:28 PM
brentwoodbc
[SOLVED] my last two double angle questions
I am pretty sure I'm almost done the first one.
#1
Use the identity for sin2x and an identity for cos2x to verify that http://www.sosmath.com/CBB/latexrend...c7327fd889.gif

but when I use the formula for double angles like where http://www.sosmath.com/CBB/latexrend...3887eaaaf8.gifwith a http://www.sosmath.com/CBB/latexrend...b5c1c816b4.gif
I think it is
http://www.sosmath.com/CBB/latexrend...451be35bae.gif

And
#2
how do you find roots/solve something like
$\displaystyle 4sinxcosx-1=0$ or $\displaystyle cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

sorry about the multiple questions but the site was down lol.
• May 8th 2009, 03:58 PM
Shyam
Quote:

Originally Posted by brentwoodbc
I am pretty sure I'm almost done the first one.
#1
Use the identity for sin2x and an identity for cos2x to verify that http://www.sosmath.com/CBB/latexrend...c7327fd889.gif

but when I use the formula for double angles like where http://www.sosmath.com/CBB/latexrend...3887eaaaf8.gifwith a http://www.sosmath.com/CBB/latexrend...b5c1c816b4.gif
I think it is
http://www.sosmath.com/CBB/latexrend...451be35bae.gif

$\displaystyle \sin ^2 2x + \cos ^2 2x \hfill \\$

$\displaystyle = \left( {2\sin x\cos x} \right)^2 + \left( {\cos ^2 x - \sin ^2 x} \right)^2 \hfill \\$

$\displaystyle = 4\sin ^2 x\cos ^2 x + \cos ^4 x - 2\cos ^2 x\sin ^2 x + \sin ^4 x \hfill \\$

$\displaystyle = \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$

$\displaystyle = \left( {\sin ^2 x + \cos ^2 x} \right)^2 = 1^2 = 1 \hfill \\$
• May 8th 2009, 04:03 PM
Shyam
Quote:

Originally Posted by brentwoodbc
#2
how do you find roots/solve something like
$\displaystyle 4sinxcosx-1=0$ or $\displaystyle cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

sorry about the multiple questions but the site was down lol.

$\displaystyle 4 \sin x \cos x - 1 = 0$

$\displaystyle 2(2\sin x \cos x) - 1 = 0$

$\displaystyle 2\sin 2x -1=0$

$\displaystyle \sin 2x =\frac{1}{2}$

$\displaystyle 2x = \frac{\pi}{6}, \frac{5\pi}{6}$

$\displaystyle x = \frac{\pi}{12}, \frac{5\pi}{12}$

Now,

$\displaystyle \cos x - sin^2 x =1$

Change $\displaystyle sin^2 x$ into $\displaystyle 1- \cos ^2 x$, then solve the quadratic eqn that is formed.
• May 8th 2009, 04:16 PM
brentwoodbc
thanks, very much.
ya on the first one factoring the
$\displaystyle = \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$
was confusing me because of all the $\displaystyle sin^4$ 's etc.